/*A strange lift
There is a strange lift.The lift can stop can at every floor as you want, and there is a number Ki(0 <= Ki <= N) on every floor.The lift have just two buttons: up and down.When you at floor i,if you press the button "UP" , you will go up Ki floor,i.e,you will go to the i+Ki th floor,as the same, if you press the button "DOWN" , you will go down Ki floor,i.e,you will go to the i-Ki th floor. Of course, the lift can't go up high than N,and can't go down lower than 1. For example, there is a buliding with 5 floors, and k1 = 3, k2 = 3,k3 = 1,k4 = 2, k5 = 5.Begining from the 1 st floor,you can press the button "UP", and you'll go up to the 4 th floor,and if you press the button "DOWN", the lift can't do it, because it can't go down to the -2 th floor,as you know ,the -2 th floor isn't exist.
Here comes the problem: when you are on floor A,and you want to go to floor B,how many times at least he has to press the button "UP" or "DOWN"?
Input
The input consists of several test cases.,Each test case contains two lines.
The first line contains three integers N ,A,B( 1 <= N,A,B <= 200) which describe above,The second line consist N integers k1,k2,....kn.
A single 0 indicate the end of the input.
Output
For each case of the input output a interger, the least times you have to press the button when you on floor A,and you want to go to floor B.If you can't reach floor B,printf "-1".
Sample Input
5 1 5
3 3 1 2 5
0
Sample Output
3
思路：从给定的点A走到点B，n是给定的楼层数量，第二行输入的是在当前的楼层可以跳转的楼层数，问最小需要多少步可以到达，注意边界的控制即可。
*/
#include <bits/stdc++.h>
const int maxn = 222;
using namespace std;

int a[maxn], n, A, B;
bool vis[maxn];

struct data
{
    int x;//当前所在节点
    int step;// 走过的步数
}st, pre;

int BFS()
{
    queue<data> que;
    st.x = A;
    st.step = 0;
    vis[A] = 1;
    que.push(st);

    while(!que.empty())
    {
        st = que.front();
        que.pop();

        if (st.x == B)
            return st.step;

        // 两个方向移动，分别判断是否可以走
        int x = st.x + a[st.x];
        if(x >= 1 && x <= n && !vis[x])
        {
            vis[x] = 1;
            pre.x = x;
            pre.step = st.step + 1;
            que.push(pre);
        }
        x = st.x - a[st.x];
        if (x >= 1 && x <= n && !vis[x])
        {
            vis[x] = 1;
            pre.x = x;
            pre.step = st.step + 1;
            que.push(pre);
        }
    }
    return -1;
}

int main()
{
    while (~scanf("%d", &n))
    {
        if (!n)
            break;
        scanf("%d%d", &A, &B);
        for (int i = 1;i <= n;i ++)
            scanf("%d", &a[i]);
        memset(vis, 0, sizeof(vis));
        printf("%d\n", BFS());
    }
    return 0;
}
